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    \begin{enumerate}
        \item[7.]
            \begin{tabular}[t]{@{}p{.5\linewidth}l}
                $y = f(x)=\frac{1}{2}x^3$ & $x=2$ and $\Delta{x}= dx = 0.1$
            \end{tabular}\\
            \begin{align*}
            y' & = f'(x) =\frac{3}{2}x^2 \\
            dy & = f'(x)dx = \frac{3}{2} \cdot 2^2 \cdot 0.1 = 0.6\\
            \Delta{y} &= f(x+\Delta{x}) - f(x)\\
            &= \left(\frac{1}{2}(2+0.1)^3 - \frac{1}{2}(2)^3 \right) = 0.6305\\
            \end{align*}
            

        \item[8.] $y = f(x)=1-2x^2$,~~~ $x=0$,~~~ $\Delta{x}= dx = -0.1$\\
            \begin{align*}
            y' & = f'(x)=-4x \\
            dy & = f'(x)dx = -4 \cdot 0 \cdot -0.1 = 0\\
            \Delta{y} & = f(x+\Delta{x}) - f(x)\\ 
            &= \left((1-2(0+-0.1)^2) - (1-2(0)^2)\right) = -0.02\\
            \end{align*}

        \item[9.] $y = f(x)=x^4-1$,~~~ $x=-1$,~~~ $\Delta{x} = dx = 0.01$\\
            \begin{align*}
            y' &= f'(x)=4x^3 \\
            dy &= f'(x)dx = 4(-1)^3 \cdot 0.01 = -0.04\\
            \Delta{y} &= f(x+\Delta{x}) - f(x)\\
            &= \left(((-1+0.01)^4 - 1) - ((-1)^4 - 1)\right) = -0.03940399\\
            \end{align*}

        \item[10.] $y = f(x)=2x+1$,~~~ $x=2$,~~~ $\Delta{x} = dx = 0.01$\\
            \begin{align*}
            y' &= f'(x)= 2 \\
            dy &= f'(x)dx = 2 \cdot 0.01 = 0.02\\
            \Delta{y} &= f(x+\Delta{x}) - f(x)\\
            &= \left( 2(2+0.01)+1 - 2\cdot2+1 \right) = 0.02\\
            \end{align*}

    \end{enumerate}

    \begin{enumerate}
        \item[11.] 
            \begin{align*}
            y &= f(x) = 3x^2 - 4 \\
            \frac{dy}{dx} &= f'(x) = 6x \\
            dy &= f'(x)~dx = 6x~dx\\
            \end{align*}

        \item[12.] 
            \begin{align*}
            y &= f(x) = 3x^{\frac{2}{3}} \\
            \frac{dy}{dx} &= f'(x) = \frac{2}{x^(1/3)}\\
            dy &= f'(x)~dx = \frac{2}{x^(1/3)}~dx\\
            \end{align*}

        \item[13.] 
            \begin{align*}
            y &= f(x) = \frac{x+1}{2x-1} \\
            \frac{dy}{dx} &= f'(x) = \frac{(2x-1)\cdot1 - (x+1)\cdot2}{(2x-1)^2} & \text{using quotient rule}~\frac{v~du - u~dv}{dv^2}\\
        	 \frac{dy}{dx} &= f'(x) = \frac{2x-1 -2x-2}{(2x-1)^2} & \text{simplifying} \\
        	\frac{dy}{dx} &= f'(x) = \frac{-3}{4 x^2-4 x+1}\\
            dy &= f'(x)~dx = \frac{-3}{4 x^2-4 x+1}~dx\\
            \end{align*}

        \item[14.] 
            \begin{align*}
            y &= f(x) = \sqrt{9-x^2} \\
            \frac{dy}{dx} &= f'(x) = -2x~\frac{(9-x^2)^{-1/2}}{2} & \text{using the chain rule}\\
            \frac{dy}{dx} &= f'(x) = -\frac{x}{\sqrt{9 - x^2}}\\
            dy &= f'(x)~dx = -\frac{x}{\sqrt{9 - x^2}}dx\\
            \end{align*}

        \item[15.] 
            \begin{align*}
            y &= f(x) = x~\sqrt{1-x^2} \\
            \frac{dy}{dx} &= f'(x) = x \cdot \frac{1}{2}(1-x^2)^{-1/2}\cdot -2x + \sqrt{1-x^2}\\
            \frac{dy}{dx} &= f'(x) = -x^2(1-x^2)^{-1/2} + \sqrt{1-x^2}\\
            \frac{dy}{dx} &= f'(x) = \frac{-x^2}{\sqrt{1-x^2}} + \sqrt{1-x^2}\\
            dy &= f'(x)~dx = \left(\frac{-x^2}{\sqrt{1-x^2}} + \sqrt{1-x^2}\right)dx\\
            \end{align*}

        \item[16.] 
            \begin{align*}
            y &= f(x) = \sqrt{x}+\frac{1}{\sqrt{x}} \\
            \frac{dy}{dx} &= f'(x) = \frac{1}{2}x^{-1/2} + \frac{-1}{2}x^{-3/2}\\
            \frac{dy}{dx} &= f'(x) = \frac{1}{2\sqrt{x}} + \frac{-1}{2x^{3/2}}\\
            dy &= f'(x)~dx = \left(\frac{1}{2\sqrt{x}} + \frac{-1}{2x^{3/2}} \right)dx\\
            \end{align*}

        \item[17.] 
            \begin{align*}
            y &= f(x) = 2x - \cot^2{x} \\
            \frac{dy}{dx} &= f'(x) = 2 - 2\cot{x}\cdot(-\csc^2{x})\\
            f'(x)~dx &= \left(2 - 2\cot{x}\cdot(-\csc^2{x})\right)dx\\
            f'(x)~dx &= \left(2 + 2\cot{x}\cdot\csc^2{x}\right)dx\\
            \end{align*}

        \item[18.] 
            \begin{align*}
            y &= f(x) = x\sin{x} \\
            \frac{dy}{dx} &= f'(x) =  \sin{x} + x\cos{x}\\
            dy &= f'(x)~dx = f'(x) =  \left( \sin{x} + x\cos{x} \right)dx\\
            \end{align*}

        \item[19.] 
            \begin{align*}
            y &= f(x) = \frac{1}{3}\cos\left({\frac{6\pi x - 1}{2}}\right) \\
            \frac{dy}{dx} &= f'(x) = \frac{1}{3} \cdot -\sin{\left(\frac{6\pi x - 1}{2}\right)} \cdot \frac{6\pi}{2}\\
            \frac{dy}{dx} &= f'(x) = \frac{-6\pi}{6} \sin{\left(\frac{6\pi x - 1}{2}\right)}\\
            dy &= f'(x)~dx = -\pi\sin{\left(\frac{6\pi x - 1}{2}\right)}dx\\
            \end{align*}

        \item[20.] 
            \begin{align*}
            y &= f(x) = \frac{\sec^2{x}}{x^2+1} \\
            \frac{dy}{dx} &= f'(x) = \frac{(x^2+1)\left(\frac{d}{dx}(\sec^2{x})\right) - \sec^2{x}\cdot 2x }{ (x^2+1)^2}\\
            \frac{dy}{dx} &= f'(x) = \frac{(x^2+1)\left(  2\sec{x}\cdot\tan{x}\sec{x}  \right) - \sec^2{x}\cdot 2x }{ (x^2+1)^2}\\
            \frac{dy}{dx} &= f'(x) = \frac{(x^2+1)\left(  2\sec^2{x}\tan{x} \right) - \sec^2{x}\cdot 2x }{ (x^2+1)^2}\\
            dy &= f'(x)~dx = \left( \frac{(x^2+1)\left(  2\sec^2{x}\tan{x} \right) - \sec^2{x}\cdot 2x }{ (x^2+1)^2} \right)dx\\
            \end{align*}
    \end{enumerate}

    \begin{enumerate}
        \item[21.] 
        (a) From the graph we know that $f(2) = f'(2) = 1$ and that $\Delta{x} = 1.9 - 2 = 0.1$, Thus:
            \begin{align*}
        	f(x + \Delta{x}) &\approx f(x) + f'(x)dx \\
        	f(2-0.1) &\approx f(2) + f'(2)(-0.1)\\
        	f(1.9) &\approx 1 + 1\cdot(-0.1) = 0.9\\
            \end{align*}
        (b) From the graph we know that $f(2) = f'(2) = 1$ and that $\Delta{x} = 2.04 - 2 = 0.04$, Thus:
            \begin{align*}
        	f(x + \Delta{x}) &\approx f(x) + f'(x)dx \\
        	f(2+0.04) &\approx f(2) + f'(2)(0.04)\\
        	f(2.04) &\approx 1 + 1\cdot(0.04) = 1.04\\
            \end{align*}
        \item[25.] 
        (a) Given that g(3) = 8 and from the graph we know that $g'(3) = -1/2$, thus:
            \begin{align*}
            \Delta{x} &= 2.93 - 3 = -0.07\\
        	g(x + \Delta{x}) &\approx g(x) + g'(x)dx \\
        	g(3-0.07) &\approx g(3) + g'(3)(-0.07)\\
        	g(2.93) &\approx 8 + \frac{-1}{2}\cdot -0.07 = 8.035\\
            \end{align*}
        (b) Given that g(3) = 8 and from the graph we know that $g'(3) = -1/2$, thus:
            \begin{align*}
            \Delta{x} &= 3.1 - 3 = 0.1\\
        	g(x + \Delta{x}) &\approx g(x) + g'(x)dx \\
        	g(3+0.1) &\approx g(3) + g'(3)(0.1)\\
        	g(3.1) &\approx 8 + \frac{-1}{2}\cdot 0.1 = 7.95\\
            \end{align*}
    \end{enumerate}

    \begin{enumerate}
        \item[43.] 
            \begin{tabular}[bc]{ccc}
            %\begin{tabular}[t]{@{}p{.5\linewidth}lll}
            $x = 100$ & and & $dx = -0.6$\\
            \end{tabular}
            \begin{align*}
            f(x + \Delta{x}) &\approx f(x) + f'(x)~dx \\
            &= \sqrt{x} + \frac{1}{2\sqrt{x}}~dx\\
            &= \sqrt{100} + \frac{1}{2\sqrt{100}}\cdot-0.6 = 9.97\\
            \end{align*}
	with a calculator: $\sqrt{99.4} = 9.96995... $


        \item[45.] 
            \begin{tabular}[bc]{ccc}
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            $x = 625$ & and & $dx = -1$\\
            \end{tabular}
            \begin{align*}
            f(x + \Delta{x}) &\approx f(x) + f'(x)~dx \\
            &= \sqrt[4]{x} + \frac{1}{4x^{3/4}}dx\\
            &= \sqrt[4]{625} + \frac{1}{4(625)^{3/4}}\cdot-1\\
            &= 5 + \frac{1}{500}\cdot-1 = 4.998\\
            \end{align*}
	with a calculator: $\sqrt[4]{624} = 4.99799... $
    \end{enumerate}



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